Solve:
$$ x^2 \left ( y -x {dy \over dx}\right ) = y \left ( dy\over dx\right)^2$$
Soln:
$$ y^2 = cx^2 + c^2 $$

Question

Solve: 
$$ x^2 \left ( y -x {dy \over dx}\right ) = y \left ( dy\over dx\right)^2$$
Soln:
$$  y^2 = cx^2 + c^2 $$

turingtest · Answer

*bookmark!

anonymous · Answer

You're doing DEs now?  Okay...  I'll help you tomorrow if no one answers your question.  But I'm tired rite now.

anonymous · Answer

multiply both sides of equation by  $y$  u will get
$$ x^2y^2-x^3y \frac{dy}{dx}= (y \frac{dy}{dx})^2 $$
suppose that $t$ is a function of $x$

let
$$ y \frac{dy}{dx}=tx $$ plug in the original equation :

$x^2y^2-tx^4= t^2x^2 $  or   $y^2=tx^2+ t^2 $  now take derivative of both sides we have
$$2 y \frac{dy}{dx}=2xt=2xt+x^2 \frac{dt}{dx}+2t\frac{dt}{dx} $$
or  
$$x^2 \frac{dt}{dx}+2t\frac{dt}{dx}=0 $$now we have 2 options  ---->  $t'=0$   or  $2t=-x^2$


case  $(I)$  ......... $t'=0$

$t=c$  or  $y\frac{dy}{dx}=cx$  and  $y^2=cx^2+d$  put back in this original equation we will get $d=c^2$

so for this case answer is  $y^2=cx^2+c^2$


case  $(II)$ ..........$2t=-x^2$

$2xt=-x^3$  or  $2y\frac{dy}{dx}=-x^3$  and from this case we get 
$$ y^2=-\frac{x^4}{4}+b$$substituting this in the original equation gives $b=0$ so  $y^2=-x^4/4$  and this is not acceptable...

anonymous · Answer

***put back this in.............lol

experimentx · Answer

this is a guessing game!!
right substitution would yield in to Clairaut's equation which isn't difficult to solve afterwards.

experimentx · Answer

the problems is how to think like the people who devise these kinds of problem ... i could never do this.

anonymous · Answer

me too...maybe i could never do this...i used the answer to get the right sub

experimentx · Answer

x^2 = u, y^2 = v for this

experimentx · Answer

i have another problem i'm having trouble with (have answer but no solution) .... i'll post in a while.